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As long as $Z$ is projective or quasi-projective over say $k = \mathbb{C}$, this is fine. This can be generalized, but let me keep it simple for now.

The quasi-projective case reduces to the projective case by taking the closure of $Z$ in projective space.
Therefore, let's do the projective case, $Z \subseteq X = \mathbb{P}^n_k$. Let $I_Z$ denote the ideal sheaf of $Z$. Consider $$I_Z \otimes O_X(n)$$ for $n \gg 0$. This sheaf is globally generated, and so has no basepoints away from $Z$. For a general section $\gamma \in \Gamma(X, I_Z \otimes O_X(n))$, the hypersurface $H = V(\gamma)$ is therefore smooth away from $Z$ (here I'm using Bertini -- characteristic zero and algebraically closed). Any of these hyperplanes passes through $Z$ by construction.

In fact, $W = Z \cup Y$ where $Y$ is some other irreducible scheme which is smooth away from from $Z$ (the fact that $Y$ is irreducible and smooth away from $W$ comes from Bertini's theorem).

Ok, how do you know that the irreducible component of $W$ corresponding to $Z$ is reduced? This is also pretty easy. This comes from the fact that $Z$ was reduced in the first place. Indeed, your general sections $\gamma_1, \dots, \gamma_d$ generate maximal ideal of the stalk at the generic point of $W$ since $I_Z \otimes O_X(n)$ was globally generated.

Of course, for arbitrary schemes, there's no hope. $Z$ need not embed in a nonsingular scheme at all.